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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 14"

(→‎Solution (Law of Cosines and Equilateral Triangle Area))
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==Problem==
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In the figure, equilateral hexagon <math>ABCDEF</math> has three nonadjacent acute interior angles that each measure <math>30^\circ</math>. The enclosed area of the hexagon is <math>6\sqrt{3}</math>. What is the perimeter of the hexagon?
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<asy>
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size(10cm);
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pen p=black+linewidth(1),q=black+linewidth(5);
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pair C=(0,0),D=(cos(pi/12),sin(pi/12)),E=rotate(150,D)*C,F=rotate(-30,E)*D,A=rotate(150,F)*E,B=rotate(-30,A)*F;
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draw(C--D--E--F--A--B--cycle,p);
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dot(A,q);
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dot(B,q);
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dot(C,q);
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dot(D,q);
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dot(E,q);
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dot(F,q);
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label("$C$",C,2*S);
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label("$D$",D,2*S);
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label("$E$",E,2*S);
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label("$F$",F,2*dir(0));
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label("$A$",A,2*N);
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label("$B$",B,2*W);
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</asy>
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<math>\textbf{(A)} 4 \qquad \textbf{(B)} 4\sqrt3 \qquad \textbf{(C)} 12 \qquad \textbf{(D)} 18 \qquad \textbf{(E)} 12\sqrt3</math>
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==Solution (Law of Cosines and Equilateral Triangle Area)==
 
==Solution (Law of Cosines and Equilateral Triangle Area)==
  
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So, the total area of the hexagon is the area of the equilateral triangle plus thrice the area of each isosceles triangle or <math>\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2+3(\frac{1}{4}s^2)=\frac{\sqrt{3}}{2}s^2=6\sqrt{3}</math>. The perimeter is <math>6s=\boxed{12\sqrt{3}}</math>.
 
So, the total area of the hexagon is the area of the equilateral triangle plus thrice the area of each isosceles triangle or <math>\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2+3(\frac{1}{4}s^2)=\frac{\sqrt{3}}{2}s^2=6\sqrt{3}</math>. The perimeter is <math>6s=\boxed{12\sqrt{3}}</math>.
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{{AMC12 box|year=2021 Fall|ab=A|num-b=13|num-a=35}}
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{{MAA Notice}}

Revision as of 19:58, 23 November 2021

Problem

In the figure, equilateral hexagon $ABCDEF$ has three nonadjacent acute interior angles that each measure $30^\circ$. The enclosed area of the hexagon is $6\sqrt{3}$. What is the perimeter of the hexagon? [asy] size(10cm); pen p=black+linewidth(1),q=black+linewidth(5); pair C=(0,0),D=(cos(pi/12),sin(pi/12)),E=rotate(150,D)*C,F=rotate(-30,E)*D,A=rotate(150,F)*E,B=rotate(-30,A)*F; draw(C--D--E--F--A--B--cycle,p); dot(A,q); dot(B,q); dot(C,q); dot(D,q); dot(E,q); dot(F,q); label("$C$",C,2*S); label("$D$",D,2*S); label("$E$",E,2*S); label("$F$",F,2*dir(0)); label("$A$",A,2*N); label("$B$",B,2*W); [/asy] $\textbf{(A)} 4 \qquad \textbf{(B)} 4\sqrt3 \qquad \textbf{(C)} 12 \qquad \textbf{(D)} 18 \qquad \textbf{(E)} 12\sqrt3$

Solution (Law of Cosines and Equilateral Triangle Area)

Isosceles triangles $ABE$, $CBD$, and $EDF$ are identical by SAS similarity. $BF=BD=DF$ by CPCTC, and triangle $BDF$ is equilateral.

Let the side length of the hexagon be $s$.

The area of each isosceles triangle is $\frac{1}{2}s^2\sin{30}=\frac{1}{4}s^2$ by the fourth formula here.

By the Law of Cosines, the square of the side length of equilateral triangle BDF is $2s^2-2x^2\cos{30}=(2-\sqrt{3})s^2$. Hence, the area of the equilateral triangle is $\frac{\sqrt{3}}{4}(2-\sqrt{3})s^2=\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2$.

So, the total area of the hexagon is the area of the equilateral triangle plus thrice the area of each isosceles triangle or $\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2+3(\frac{1}{4}s^2)=\frac{\sqrt{3}}{2}s^2=6\sqrt{3}$. The perimeter is $6s=\boxed{12\sqrt{3}}$.

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 35
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All AMC 12 Problems and Solutions

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